3.25.57 \(\int \frac {96 x^2+24 x^3+3 x^5+(32+8 x+x^3) \log (2)+(32+16 x+2 x^2-2 x^3) \log (2) \log (\frac {16+8 x+x^2-x^3}{x^2})}{(-48 x^3-24 x^4-3 x^5+3 x^6+(-16 x-8 x^2-x^3+x^4) \log (2)) \log (\frac {16+8 x+x^2-x^3}{x^2})} \, dx\)

Optimal. Leaf size=25 \[ \log \left (3+\frac {\log (2)}{x^2}\right )+\log \left (\log \left (\left (1+\frac {4}{x}\right )^2-x\right )\right ) \]

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Rubi [A]  time = 0.85, antiderivative size = 47, normalized size of antiderivative = 1.88, number of steps used = 8, number of rules used = 7, integrand size = 127, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {6688, 6742, 266, 36, 29, 31, 6684} \begin {gather*} \frac {\log (4) \log \left (3 x^2+\log (2)\right )}{2 \log (2)}+\log \left (\log \left (\frac {16}{x^2}-x+\frac {8}{x}+1\right )\right )-\frac {\log (4) \log (x)}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(96*x^2 + 24*x^3 + 3*x^5 + (32 + 8*x + x^3)*Log[2] + (32 + 16*x + 2*x^2 - 2*x^3)*Log[2]*Log[(16 + 8*x + x^
2 - x^3)/x^2])/((-48*x^3 - 24*x^4 - 3*x^5 + 3*x^6 + (-16*x - 8*x^2 - x^3 + x^4)*Log[2])*Log[(16 + 8*x + x^2 -
x^3)/x^2]),x]

[Out]

-((Log[4]*Log[x])/Log[2]) + (Log[4]*Log[3*x^2 + Log[2]])/(2*Log[2]) + Log[Log[1 + 16/x^2 + 8/x - x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\frac {2 \left (16+8 x+x^2-x^3\right ) \log (2)}{3 x^2+\log (2)}-\frac {32+8 x+x^3}{\log \left (1+\frac {16}{x^2}+\frac {8}{x}-x\right )}}{x \left (16+8 x+x^2-x^3\right )} \, dx\\ &=\int \left (-\frac {\log (4)}{x \left (3 x^2+\log (2)\right )}+\frac {32+8 x+x^3}{(-4+x) x \left (4+3 x+x^2\right ) \log \left (1+\frac {16}{x^2}+\frac {8}{x}-x\right )}\right ) \, dx\\ &=-\left (\log (4) \int \frac {1}{x \left (3 x^2+\log (2)\right )} \, dx\right )+\int \frac {32+8 x+x^3}{(-4+x) x \left (4+3 x+x^2\right ) \log \left (1+\frac {16}{x^2}+\frac {8}{x}-x\right )} \, dx\\ &=\log \left (\log \left (1+\frac {16}{x^2}+\frac {8}{x}-x\right )\right )-\frac {1}{2} \log (4) \operatorname {Subst}\left (\int \frac {1}{x (3 x+\log (2))} \, dx,x,x^2\right )\\ &=\log \left (\log \left (1+\frac {16}{x^2}+\frac {8}{x}-x\right )\right )-\frac {\log (4) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 \log (2)}+\frac {(3 \log (4)) \operatorname {Subst}\left (\int \frac {1}{3 x+\log (2)} \, dx,x,x^2\right )}{2 \log (2)}\\ &=-\frac {\log (4) \log (x)}{\log (2)}+\frac {\log (4) \log \left (3 x^2+\log (2)\right )}{2 \log (2)}+\log \left (\log \left (1+\frac {16}{x^2}+\frac {8}{x}-x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 31, normalized size = 1.24 \begin {gather*} -2 \log (x)+\log \left (3 x^2+\log (2)\right )+\log \left (\log \left (1+\frac {16}{x^2}+\frac {8}{x}-x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(96*x^2 + 24*x^3 + 3*x^5 + (32 + 8*x + x^3)*Log[2] + (32 + 16*x + 2*x^2 - 2*x^3)*Log[2]*Log[(16 + 8*
x + x^2 - x^3)/x^2])/((-48*x^3 - 24*x^4 - 3*x^5 + 3*x^6 + (-16*x - 8*x^2 - x^3 + x^4)*Log[2])*Log[(16 + 8*x +
x^2 - x^3)/x^2]),x]

[Out]

-2*Log[x] + Log[3*x^2 + Log[2]] + Log[Log[1 + 16/x^2 + 8/x - x]]

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fricas [A]  time = 0.63, size = 34, normalized size = 1.36 \begin {gather*} \log \left (3 \, x^{2} + \log \relax (2)\right ) - 2 \, \log \relax (x) + \log \left (\log \left (-\frac {x^{3} - x^{2} - 8 \, x - 16}{x^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+2*x^2+16*x+32)*log(2)*log((-x^3+x^2+8*x+16)/x^2)+(x^3+8*x+32)*log(2)+3*x^5+24*x^3+96*x^2)/(
(x^4-x^3-8*x^2-16*x)*log(2)+3*x^6-3*x^5-24*x^4-48*x^3)/log((-x^3+x^2+8*x+16)/x^2),x, algorithm="fricas")

[Out]

log(3*x^2 + log(2)) - 2*log(x) + log(log(-(x^3 - x^2 - 8*x - 16)/x^2))

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giac [A]  time = 0.31, size = 36, normalized size = 1.44 \begin {gather*} \log \left (3 \, x^{2} + \log \relax (2)\right ) - 2 \, \log \relax (x) + \log \left (-\log \left (-x^{3} + x^{2} + 8 \, x + 16\right ) + \log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+2*x^2+16*x+32)*log(2)*log((-x^3+x^2+8*x+16)/x^2)+(x^3+8*x+32)*log(2)+3*x^5+24*x^3+96*x^2)/(
(x^4-x^3-8*x^2-16*x)*log(2)+3*x^6-3*x^5-24*x^4-48*x^3)/log((-x^3+x^2+8*x+16)/x^2),x, algorithm="giac")

[Out]

log(3*x^2 + log(2)) - 2*log(x) + log(-log(-x^3 + x^2 + 8*x + 16) + log(x^2))

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maple [A]  time = 0.23, size = 34, normalized size = 1.36

method	result	size
default	\(\ln \left (\ln \left (\frac {-x^{3}+x^{2}+8 x +16}{x^{2}}\right )\right )-2 \ln \relax (x )+\ln \left (3 x^{2}+\ln \relax (2)\right )\)	\(34\)
norman	\(\ln \left (\ln \left (\frac {-x^{3}+x^{2}+8 x +16}{x^{2}}\right )\right )-2 \ln \relax (x )+\ln \left (3 x^{2}+\ln \relax (2)\right )\)	\(34\)
risch	\(\ln \left (\ln \left (\frac {-x^{3}+x^{2}+8 x +16}{x^{2}}\right )\right )-2 \ln \relax (x )+\ln \left (3 x^{2}+\ln \relax (2)\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+2*x^2+16*x+32)*ln(2)*ln((-x^3+x^2+8*x+16)/x^2)+(x^3+8*x+32)*ln(2)+3*x^5+24*x^3+96*x^2)/((x^4-x^3-
8*x^2-16*x)*ln(2)+3*x^6-3*x^5-24*x^4-48*x^3)/ln((-x^3+x^2+8*x+16)/x^2),x,method=_RETURNVERBOSE)

[Out]

ln(ln((-x^3+x^2+8*x+16)/x^2))-2*ln(x)+ln(3*x^2+ln(2))

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maxima [A]  time = 0.64, size = 35, normalized size = 1.40 \begin {gather*} \log \left (3 \, x^{2} + \log \relax (2)\right ) - 2 \, \log \relax (x) + \log \left (\log \left (x^{2} + 3 \, x + 4\right ) - 2 \, \log \relax (x) + \log \left (-x + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+2*x^2+16*x+32)*log(2)*log((-x^3+x^2+8*x+16)/x^2)+(x^3+8*x+32)*log(2)+3*x^5+24*x^3+96*x^2)/(
(x^4-x^3-8*x^2-16*x)*log(2)+3*x^6-3*x^5-24*x^4-48*x^3)/log((-x^3+x^2+8*x+16)/x^2),x, algorithm="maxima")

[Out]

log(3*x^2 + log(2)) - 2*log(x) + log(log(x^2 + 3*x + 4) - 2*log(x) + log(-x + 4))

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mupad [B]  time = 2.15, size = 33, normalized size = 1.32 \begin {gather*} \ln \left (\ln \left (\frac {-x^3+x^2+8\,x+16}{x^2}\right )\right )+\ln \left (x^2+\frac {\ln \relax (2)}{3}\right )-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(96*x^2 + 24*x^3 + 3*x^5 + log(2)*(8*x + x^3 + 32) + log(2)*log((8*x + x^2 - x^3 + 16)/x^2)*(16*x + 2*x^2
- 2*x^3 + 32))/(log((8*x + x^2 - x^3 + 16)/x^2)*(log(2)*(16*x + 8*x^2 + x^3 - x^4) + 48*x^3 + 24*x^4 + 3*x^5
- 3*x^6)),x)

[Out]

log(log((8*x + x^2 - x^3 + 16)/x^2)) + log(log(2)/3 + x^2) - 2*log(x)

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sympy [B]  time = 0.53, size = 44, normalized size = 1.76 \begin {gather*} - 2 \left (\frac {\log {\relax (x )}}{\log {\relax (2 )}} - \frac {\log {\left (x^{2} + \frac {\log {\relax (2 )}}{3} \right )}}{2 \log {\relax (2 )}}\right ) \log {\relax (2 )} + \log {\left (\log {\left (\frac {- x^{3} + x^{2} + 8 x + 16}{x^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+2*x**2+16*x+32)*ln(2)*ln((-x**3+x**2+8*x+16)/x**2)+(x**3+8*x+32)*ln(2)+3*x**5+24*x**3+96*x
**2)/((x**4-x**3-8*x**2-16*x)*ln(2)+3*x**6-3*x**5-24*x**4-48*x**3)/ln((-x**3+x**2+8*x+16)/x**2),x)

[Out]

-2*(log(x)/log(2) - log(x**2 + log(2)/3)/(2*log(2)))*log(2) + log(log((-x**3 + x**2 + 8*x + 16)/x**2))

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